Momentum Principle
Energy
Angular Momentum
Springs
Fundamental Interactions
100
Which of these would NOT be an example of an inelastic collision?
A. Two cars crash into each other, and stop with a loud bang
B. A match scrapes a matchbook and bursts into flame
C. A man is clapping his hands such that they move with equal, but opposite velocities
D. Two hydrogen atoms fuse together to form a helium atom and gamma radiation
E. Neutrons fuse with hydrogen atoms in a nuclear reactor core such that kinetic energy is conserved
E. Neutrons fuse with hydrogen atoms in a nuclear reactor core such that kinetic energy is conserved
The difference between an elastic and an inelastic collision is the loss or conservation of kinetic energy. In an inelastic collision kinetic energy is not conserved, and will change forms into sound, heat, radiation, or some other form. In an elastic collision kinetic energy is conserved and does not change forms.
Remember, total energy and total momentum are conserved regardless of the type of collision; however, while energy cannot be created nor destroyed, it can change forms.
In the answer options, only one choice preserves the total kinetic energy. The resulting bang from the car crash, the flame from the match, the sound of hands clapping, and the gamma radiation during hydrogen fusion are all examples of the conversion of kinetic energy to other forms, making each of these an inelastic collision. Only the neutron fusion described maintains the conservation of kinetic energy, making this an elastic collision.
100
A ball drops from a height h. What more do we need to calculate initial potential energy?
A. Final velocity
B. Elasticity of the ball
C. Mass
D. Horizontal displacement
E. Initial velocity
Mass
Explanation:
The formula for potential energy is:
PE=mgh
Since g is a constant, the acceleration due to gravity on Earth, we only need the mass and the height. The problem gives a height, so we only need mass.
Given the mass and the height, we would be able to calculate the initial potential energy.
100
A 2kg ball rolls around the edge of a circle with a radius of 1m. If it is rolling at a speed of 3ms, what is the period of the ball?
A. π s
B. 2π/3 s
C. 3π/2 s
D. π/2 s
E. 2π s
2π/3 s
Explanation:
The period, T, of an object moving in circular motion is the amount of time it takes for the object to make one complete loop of the circle.
If we start with the linear understanding of velocity, v=ΔxΔt, we can apply the same concept here. Our velocity should be the change in distance over the change in time. In this case, we don't have a definite time, t, but we do have a period in terms of one complete loop.
We can set up an equation for the period using the circumference of the circle as our distance: v=2πr/T.
Plug in the given values to solve for the period.
v=2πr/T
3m/s=2π(1m)/T
T=2π(1m)/3m/s
T=2π/3 s
100
What does the negative sign indicate in the equation below?
F=−kx
A. The force, F, acts downwards
B. The force, F, acts in the opposite direction of the displacement, x
C. The spring constant, k, is always negative
D. The force, F, acts in a negative direction
B. The force, F, acts in the opposite direction of the displacement, x
Explanation:
The correct answer is that the force acts in the opposite direction of the displacement.
The equation given is Hooke's law, which is used to determine spring force based on the spring constant, k, and the displacement of the spring, x. Negative and downward are arbitrary, and spring constants are always positive. If a spring is stretched or compressed in a given direction, the force of the spring will always act opposite the direction of the displacement in order to return the spring to the resting equilibrium position.
100
A 0.05kg ball falls off a cliff. What is the force of gravity on the ball? Assume g=−9.8m/s2
A. We need to know the time the ball is in the air in order to solve
B. −0.49N
C. 49N
D. 0.49N
E. We need to know the height of the cliff in order to solve
−0.49N
Explanation:
Newton's second law states:
F=ma
In this case the acceleration will be the constant acceleration due to gravity on Earth.
F=mg
Use the acceleration of gravity and the mass of the ball to solve for the force on the ball.
F=(0.05kg)(−9.8m/s2)
F=−0.49N
The answer is negative because the force is directed downward. Since gravity is always acting downward, a force due to gravity will always be negative.
200
A man with a mass of m is painting a house. He stands on a tall ladder of height h. He leans over and falls straight down off the ladder. If he is in the air for s seconds, what will be his momentum right before he hits the ground?
A. (mv)(s)
B. mgs
C. 12mv2
D. mgh
E. (mg)(s)
E. (mg)(s)
Explanation:
The problem tells us he falls vertically off the ladder (straight down), so we don't need to worry about motion in the horizontal direction.
The equation for momentum is:
p=mv
We can assume he falls from rest, which allows us to find the initial momentum.
m∗0ms=0kg∗ms.
From here, we can use the formula for impulse:
Δp=FΔt
(pf−pi)=FΔt
We know his initial momentum is zero, so we can remove this variable from the equation.
pf=FΔt
The problem tells us that his change in time is s seconds, so we can insert this in place of the time.
pf=(F)(s)
The only force acting upon man is the force due to gravity, which will always be given by the equation FG=mg.
pf=(FG)(s)
pf=(mg)(s)
200
An astronaut is on a new planet. She discovers that if she drops a space rock from 10m above the ground, it has a final velocity of 3ms just before it strikes the planet surface. What is the acceleration due to gravity on the planet?
A. −1.33m/s2
B. 0.12m/s2
C. −0.45m/s2
D. −8.88m/s2
E. −9.8m/s2
C. −0.45m/s2
Explanation:
We can use conservation of energy to solve. The potential energy when the astronaut is holding the rock will be equal to the kinetic energy just before it strikes the surface.
PE=KE
mgh=12mv2
Now, we need to solve for g, the gravity on the new planet. The masses will cancel out.
gh=12v2
Plug in the given values and solve.
g(−10m)=12(3m/s)2
g(−10m)=12(9m2/s2)
g(−10m)=(4.5m2/s2)
g=4.5m2/s2−10m
g=−0.45m/s2
200
Ellen is swinging a 0.01kg yo-yo in a circular path perpendicular to the ground. The yo-yo moves in a clockwise direction with a constant speed of 2ms.
What is the velocity of the yo-yo at the bottom of the circle?
A. 0m/s
B. 2m/s to the left
C. 2m/s to the right
D. 2m/s down
E. 2m/s up
B. 2m/s to the left
Explanation:
Remember, when working with circular motion, the velocity is ALWAYS tangential to the circle. This means that, even though the speed is constant, the direction is always tangent to the edge of the circle. If the circle below represents the path of the yo-yo, and it moves in a clockwise direction, then the velocity at the bottom of the path will be to the left.
The magnitude of the velocity is constant, so the final answer will be 2m/s to the left.
200
A spring with a spring constant of 1500Nm is compressed 0.87m. How much potential energy has been generated?
A. 1305J
B. 862.07J
C. 431.03J
D. 652.5J
E. 567.68J
E. 567.68J
Explanation:
The formula for the potential energy in a spring is:
PE=1/2kx2
Use the given spring constant and displacement to solve for the stored energy.
PE=1/2(1500Nm)(0.87m)2
PE=(750Nm)(0.7569m2)
PE=567.68J
200
Susan is trying to push a 12kg crate across the floor. She observes that the force of friction between the crate and the floor is −50N. What is the coefficient of static friction?
A. 0.23
B. 0.56
C. 0.43
D. It cannot be determined
E. 0.11
C. 0.43
Explanation:
The equation for the force of friction is Ffriction=μFnormal, where μ is the coefficient of static friction.
The normal force is equal to the mass times acceleration due to gravity, but in the opposite direction (negative of the force of gravity).
Fnormal=−Fgravity=−(mg)
Fnormal=−((12kg)(−9.8m/s2))
Fnormal=−(−117.6N)
Fnormal=117.6N
Since the problem tells us that the force due to friction is −50N, we can plug these values into our original equation to solve for the coefficient of friction.
Ffriction=μFnormal
(−50N)=μ(117.6N)
−50N/117.6N=μ
0.43=μ
The coefficient of friction has no units.
300
A space vehicle, in a circular orbit around Earth, collides with a small asteroid that ends up in the vehicle’s storage bay. For this collision
A. Only momentum is conserved
B. Only kinetic energy is conserved
C. Both momentum and kinetic energy are conserved
D. Neither momentum nor kinetic energy is conserved
Only momentum is conserved
Explanation:
This is an inelastic collision as the two objects stick together and move together with the same velocity. Inelastic collisions conserve momentum, but they do not conserve kinetic energy.
300
A skier starts at the top of a hill with 350J of energy. Assuming energy is conserved, what is her final kinetic energy?
Insufficient information to solve.
A. 26.46J
B. 350J
C. 700J
D. 175J
350J
Explanation:
If energy is conserved, then the total energy at the beginning equals the total energy at the end.
Since we have ONLY potential energy at the beginning and ONLY kinetic energy at the end, PEi=KEf.
Therefore, since our PEi=350J, our kinetic energy will also equal 350J.
300
A merry-go-round has a mass of 1500kg and radius of 8m. How much net work is required to accelerate it from rest to a ration rate of 1 revolution per 8 seconds? Assume it is a solid cylinder.
A. 14800J
B. 18000J
C. 380J
D. 8200J
E. 12800J
14800J
Explanation:
We know that the work-kinetic energy theorem states that the work done is equal to the change of kinetic energy. In rotational terms this means that
W=δKE
W=12Iω2f−1/2Iω2i
In this case the initial angular velocity is 0rad/sec.
We can convert our final angular velocity to radians per second.
1 rev/8s ∗ 2πrad/1rev = π/4 rad/s
We also can calculate the moment of inertia of the merry-go-round assuming that it is a uniform solid disk.
I=1/2mr2
I=1/2(1500kg)(8m)2
I=48000kg/m2
We can put this into our work equation now.
W=12(48000kg/m2)(π/4 rad/s)2−0
W=14800J
300
A 41N force is used to stretch a spring 0.11m. What is the spring constant?
A. 3388.43N/m
B. 0.223N/m
C. 372.73N/m
D. 0.496N/m
E. 4.51N/m
C. 372.73N/m
Explanation:
The formula for the force required to stretch or compress a spring is:
F=kΔx
We are given the force and the distance, allowing us to solve for the spring constant.
41N = k ∗ 0.11m
41N/0.11m =k
372.73N/m = k
300
A box sits on a ramp at an incline of 30 degrees. If the box has a weight of 50N, what is the normal force experienced by the box?
A. 43N
B. 424N
C. 46N
D. 490N
A. 43N
Explanation:
The correct answer is 43N. Since the box is on an incline, normal force balances with the component of gravity that is perpendicular to the surface of the incline.
FN=−Fgy=−(mgcosθ)
Note that the normal force is in the upward (positive) direction, while gravitational acceleration and the force of gravity are in the downward (negative) direction.
We are given the weight of the box (note that it will be downward and, therefore, negative):
mg=−50N
Use this to solve for the normal force.
FN=−(−50N×cos30o)
FN=43N
400
A 6kg ball is thrown west at 20ms and collides with a 14kg ball while in the air. If the balls stick together in the crash and fall straight down to the ground, what was the velocity of the second ball?
A. 9.8m/s west
B. 8.6m/s east
C. 2.3m/s east
D. 11.2m/s downward
E. 4.7m/s east
8.6ms east
Explanation:
We know that if the balls fell straight down after the crash, then the total momentum in the horizontal direction is zero. The only motion is due to gravity, rather than any remaining horizontal momentum. Based on conservation of momentum, the initial and final momentum values must be equal. If the final horizontal momentum is zero, then the initial horizontal momentum must also be zero.
pi=pf
m1v1+m2v2=m1v3+m2v4
In our situation, the final momentum is going to be zero.
m1v1+m2v2=0ms
Use the given values for the mass of each ball and initial velocity of the first ball to find the initial velocity of the second.
(6kg)(20ms)+(14kg)v2=0ms
(120N)+(14kg)(v2)=0ms
(14kg)(v2)=−120N
v2=−120N14kg=−8.6ms
The negative sign tells us the second ball is traveling in the opposite direction as the first, meaning it must be moving east.
400
A 2kg mass attached to the end of a spring causes it to stretch 5.0cm. If another 2kg mass is added to the spring, the potential energy of the spring will be
A. 3 times as much
B. Twice as much
C. 4 times as much
D. The same
Twice as much
Explanation:
Hooke’s law states that spring constant is directly related to the force applied and the distance that the object is stretched.
k=F/Δx
We also know that the potential energy of a spring is related to the spring constant and the distance that the object is stretched
PEspring=12kΔx2
We can substitute our equation for Hooke’s law into the potential energy equation.
PEspring=12(F/Δx)Δx2
This simplifies to
PEspring=12FΔx
This equation shows that there is a direct relationship between the force on the spring and the potential energy of the spring. If the force is doubled, then the potential energy will likewise double.
400
A 12N force is applied perpendicularly to a 0.4m wrench. How much torque is generated?
A. 30N∗m
B. 48N∗m
C. 1.92N∗m
D. 4.8N∗m
E. 3N∗m
4.8 N∗m
Explanation:
The formula for torque is:
τ=F∗sin(θ)∗r
In this formula, θ is the angle the force makes with the lever arm. Since our force is applied perpendicularly, this angle will be 90o. Use this angle, the force applied, and the length of the lever arm to calculate the torque.
τ=F∗sin(θ)∗r
τ=(12N)(sin(90o))(0.4m)
τ=(12N)(1)(0.4m)
τ=4.8N∗m
400
A spring with a spring constant of 1500N/m is compressed 0.87m. A 0.67kg object is placed at the end of the compressed spring and the spring is released. What is the maximum velocity of the object?
A. 13.79m/s
B. 41.17m/s
C. 190.79m/s
D. There is insufficient information to solve
E. 1694.6m/s
B. 41.17m/s
Explanation:
For this problem, use the law of conservation of energy. Assuming no other forces are acting upon the object the initial spring potential energy will be equal to the maximum final kinetic energy.
PEspring=KEf
Expand this equality with the formulas for each type of energy.
12kx2=12mv2
We are given the spring constant and displacement, allowing us to complete the left side of the equation. We are also able to plug in the mass to the left side of the equation.
1/2(1500Nm)(0.87m)2 = 12(0.67kg)v2
Solve to isolate and solve for the velocity variable.
(750Nm)(0.7569m2) = (0.335kg)v2
567.68J = (0.335kg)v2
567.68J0.335kg=v2
1694.57m2/s2=v2
sqrt( 1694.57m2/s2 )=sqrt( v2 )
41.17m/s = v
400
Two asteroids, one with a mass of 7.12∗1018kg and the other with mass 5.33∗108kg, are 10∗1010m apart. What is the gravitational force on the LARGER asteroid?
A. 4.74∗10−6N
B. 3.79∗105N
C. 3.55∗10−6N
D. 4.61∗10−10N
E. 2.53∗10−5N
E. 2.53∗10−5N
Explanation:
To solve this problem, use Newton's law of universal gravitation:
FG=G(m1m2/r2)
We are given the constant, as well as the asteroid masses and distance (radius). Using these values we can solve for the force.
FG=(6.67∗10−11m3/kg⋅s2) ((7.12∗1018kg)(5.33∗108kg)/(10∗1010m)2)
FG = (6.67∗10−11m3/kg⋅s2) (3.79∗1027kg / 10∗1021m2)
FG=(6.67∗10−11m3/kg⋅s2) (3.79∗105kg2/m2)
FG = 2.53∗10−5N
It actually doesn't matter which asteroid we're looking at; the gravitational force will be the same. This makes sense because Newton's 3rd law states that the force one asteroid exerts on the other is equal in magnitude, but opposite in direction, to the force the other asteroid exerts on it.
500
A 0.145kg baseball pitched horizontally at 27m/s strikes a bat and pops straight up to a height of 31.5m. If the contact time between the bat and ball is 2.5ms, calculate the average force between the ball and bat during contact.
A. 2436N
B. 1566N
C. 1215N
D. 2126N
E. 1438N
2126N
Explanation:
To figure this out we needed to consider the change in momentum of the ball.
We know that the impulse of the ball is equal to the change in momentum.
I=δp
The impulse is equal to the force times the time the force is applied.
I=Ft
The change in momentum is equal to the mass times the change in velocity.
δp=mδv
Therefore we can combine these equations to say
Ft=mδv
So let us look at what is happening in the x-direction.
F(2.5x10−3s)=(0.145kg)(0m/s − 27m/s)
Rearrange and solve the force in the x-direction.
F=−1566N
Next, let us determine what is happening in the y-direction. We will need to figure out the initial velocity of the ball in the y-direction using the height. At the peak, we also know the velocity is 0m/s. We also know that the acceleration due to gravity is −9.8m/s/s.
v2f=v2i+2aδx
(0m/s)2=v2i+2(−9.8m/s/s)(31.5m)
0=v2i−617.4
617.4=v2i
24.8m/s=v2i
Once we have the initial velocity, we can now determine the force in the y-direction.
F(2.5x10−3s)=(0.145kg)(24.8m/s − 0m/s)
F=1438.4N
Now that we have the force in both the x and y direction we can determine the overall resultant force using the Pythagorean Theorem.
(−1566N)2+(1438.4N)2=F2
F=2126N
500
A 0.5kg ball rolls down a 1m hill with an initial velocity of 2.45m/s. What is its maximum velocity?
A. 10.12m/s
B. 5.06m/s
C. 25.60m/s
D. 6.06m/s
E. 12.80m/s
B. 5.06m/s
Explanation:
For this problem, the ball starts with both potential and kinetic energy. The point of maximum velocity will have no potential energy. We can solve setting the initial energy and final energy equal, due to conservation of energy.
PEi + KEi = PEf + KEf
mghi + 1/2mv2i = mghf + 1/2mv2f
The masses will cancel out from all of the terms.
ghi + 1/2v2i = ghf + 12v2f
Plug in the given values and solve for the final velocity. Remember, when the ball is on the ground it has a height of zero.
(9.8m/s2)(1m) + 1/2(2.45m/s)2 = (9.8m/s2)(0m) + 1/2v2f
9.8m2/s2 + 12(6.0025m2/s2) = 0+1/2v2f
12.80125m2/s2 = 1/2v2f
25.6025m2/s2=v2f
sqrt( 25.6025m2/s2 ) = sqrt( v2f )
5.06m/s = vf
500
A heavy boy and a light girl are balanced on a massless seesaw. If they both move forward so that they are one-half their original distance from the pivot point, what will happen to the seesaw?
A. The side the girl is setting on will tilt downward
B. Nothing, the seesaw will still be balanced
C. It is impossible to say without knowing the masses and the distances
D. The side the boy is sitting on will tilt downward
B. Nothing, the seesaw will still be balanced
Explanation:
Torque is equal to the force applied perpendicular to a surface, multiplied by the radius or distance to the pivot point.
τ=FperpendicularR
In this example the boy of mass M is a distance R away and is balancing a girl of mass m at a distance r away.
MgR = mgr
If both of these kids move to a distance that is one half their original distance.
MgR2 = mgr2
The half cancels out of the equation and therefore the boy and girl will still be balanced.
500
A vertical spring with a spring constant of 11.1Nm is stationary. An 8kg mass is attached to the end of the spring. What is the maximum displacement that the spring will stretch?
A. −14.13m
B. −1.11m
C. −0.45m
D. −17.92m
E. −2.82m
A. −14.13m
Explanation:
The best way to solve this problem is by using energy. Notice that the spring on its own is stationary. That means its initial total energy at that moment is zero. When the mass is attached, the spring stretches out, giving it spring potential energy (PEspring).
Where does that energy come from? The only place it can come from is the addition of the mass. Since the system is vertical, this mass will have gravitational potential energy.
Use the law of conservation of energy to set these two energies equal to each other:
PEmass=PEspring
mgΔy=1/2kΔy2
We are trying to solve for displacement, and now we have an equation in terms of our variable.
Start by diving both sides by Δy to get rid of the Δy2 on the right side of the equation.
mgΔy/Δy=1/2kΔy2/Δy
mg=1/2kΔy
We are given values for the spring constant, the mass, and gravity. Using these values will allow use to solve for the displacement.
mg=1/2kΔy
8kg ∗ −9.8m/s2 = 12 ∗ 11.1Nm ∗ Δy
−78.4N = 5.55N/m∗ Δy
−78.4N/5.55N/m = Δy
−14.13m = Δy
Note that the displacement will be negative because the spring is stretched in the downward direction due to gravity.
500
Particles of charges +70μC, +45μC, and −100μC are placed in a line. The center charge is +45μC and is 0.40m away from each of the others. Calculate the net force on the center charge from the other two. 1μC is 1x10-6μC.
A. 253N
B. 76N
C. −76N
D. −253N
E. 177N
−76N
Explanation:We can calculate the force from each of the two charges on the center charge using Coulomb’s Law
F=k (q1q2/d2)
The first charge on the middle charge
F=(9x109)(70x10−6)(45x10−6)/(0.4m)2
F=177N
The last charge on the middle charge
F=(9x109)(45x10−6)(−100x10−6)/(0.4m)2
F=−253N
To find the total force we need to add both of these forces together.
F=177N+(−253N)
F=−76N